CTAI Teacher Handbook Class 4

Computational
Thinking

Class 4

Teacher Handbook

CENTRAL BOARD OF SECONDARY EDUCATION

First Edition: March, 2026

Country of Publication: India
Published by: Central Board of Secondary Education, Integrated Office, Sector 23, Dwarka, New
Delhi-110077

Page 1

PREFACE

The National Education Policy (NEP) aims to position India as a leader in emerging
knowledge fields by integrating technologies like AI, Machine Learning, Big Data, and
Computational Thinking into school education. It promotes technology-enabled,
interactive, and gamified learning using tools such as Augmented Reality (AR), Virtual
Reality (VR), and virtual labs to foster creativity, problem-solving, and interdisciplinary
exploration. NCFSE 23 carries this recommendation further for implementation.

While Artificial Intelligence (AI) is an important requirement, Computational Thinking
(CT) should be a broader skill, developing a foundation for learning AI. It can cover
various aspects like Cybersecurity, basic network, etc. Hence, CBSE approaches this
by integrating Computational Thinking with AI and other technological advancements,
without dependence on any platform.

The book focuses on strengthening logical reasoning through structured visual,
numerical, and real-life problems involving pattern progression, symmetry,
transformations, ordering, and multi-step thinking. Learners engage with tasks that
require comparing possibilities, identifying rules, and breaking problems into smaller
parts. The document also outlines age-appropriate pedagogy, learning resources,
assessment support, and classroom implementation guidelines to promote
competency-based, experiential learning aligned with NEP 2020.

Team CBSE

Page 2

ACKNOWLEDGEMENTS

Expert Committee

1. Dr. Karthik Raman, Core Leadership, IIT Madras Bodhan AI Foundation; Professor,
Department of Data Science and AI, Wadhwani School of Data Science and AI, IIT
Madras
2. Dr. Rajesh Kumar, Professor, Department of Electrical Engineering, MNIT, Jaipur
3. Dr. Seema Verma, Professor (ECE), Additional Project Director (Siemens CoE), NITTTR,
Bhopal
4. Dr. S. Neethi, Core Leadership, IIT Madras Bodhan AI Foundation; Professor of Practice,
Department of Data Science and AI, Wadhwani School of Data Science and AI, IIT Madras
5. Dr. Arun L. Naik, Associate Professor, Mathematics Education, Azim Premji University,
Bengaluru
6. Dr. Aanchal Chomal, Associate Professor, School of Continuing Education and
University Resource Centre, Azim Premji University
7. Dr. Ankit Vijayvargiya, Assistant Professor, School of Technology, Dhirubhai Ambani
University, Gandhinagar
8. Sh. R P Singh, Associate Prof &Additional Director, CBSE
9. Mr. Mikin Lala, (IIT Roorkee and IIM Calcutta Alumnus), Field Expert
10. Ms. Rekha Malhotra, (Alumna of British Institute (Business Management and
Advertising), Frameworks Mumbai (Master’s in Computer Science), and Workstation (3D
Animation)), Field Expert

Material Production Group

1. Mr. Jay Thakkar, Senior Technical Officer, Centre for Creative Learning, IIT Gandhinagar
2. Mr. Chris John, Project Scientist, Centre for Creative Learning, IIT Gandhinagar
3. Mr. Mikin Lala, (IIT Roorkee and IIM Calcutta Alumnus), Field Expert
4. Ms. Amatullah Mustafa Neemuchwala, Field Expert
5. Ms. Telidevara Sree Lasya, Field Expert
6. Mr. Parth Oza, Field Expert
7. Mr. Deep Mayekar, Field Expert
8. Mr. Mayank Patil, Field Expert
9. Mr. Shivraj Ugale, Field Expert
10. Mr. Raj Dhorade, Field Expert
11. Mr. Nilesh Vijay Rajput, Field Expert

The efforts of Prof. Manish Jain and his whole team from Centre for Creative Learning,
IIT Gandhinagar are also acknowledged and deeply appreciated.

Page 3

TABLE OF CONTENTS

SR. NO. CHAPTER PAGE NO.

Introduction 5
1.
How to Use This Book? 8
2.
Shapes Around Us 9
3.
Hide and Seek 16
4.
Patterns Around Us 24
5.
Thousands Around Us 35
6.
Sharing and Measuring 45
7.
Measuring Length 51
8.
The Cleanest Village 57
9.
Weigh It, Pour It 64
10.
Equal Groups 70
11.
Elephants, Tigers, and Leopards 75
12.
Fun with Symmetry 81
13.
Ticking Clocks and Turning Calendar 91
14.
The Transport Museum 97
15.
Data Handling 103
16.

Page 4

Introduction
Computational Thinking (CT) is a problem-solving approach that comprises Decomposition, Pattern
Recognition, Abstraction, Algorithm Design, Data Analysis and Troubleshooting. Computational
Thinking Skills involve solving complex problems that promote thinking skills such as critical &
creative thinking, abstraction and pattern recognition, as well as algorithmic thinking. Problem
identification and problem solving necessitate application of multidisciplinary understanding for
creating effective solutions.

Artificial intelligence (AI) is a cutting-edge technology that empowers machines and computers to
perform tasks that usually require mimicking human intelligence. These machines can perform
complex thinking processes such as data analysis, pattern recognition, prediction of trends, solving
problems and decision making. Thus, AI involves simulating cognitive processes associated with
human intelligence and is widely applicable in various sectors such as banking, healthcare, defense,
education, entertainment, agriculture and others for processing information, solving intricate
problems and for planning.

The National Education Policy (NEP) aims for India to emerge as a global leader in new emerging
knowledge domains such as artificial intelligence, machine learning, data analytics, 3 -D machining
etc. To realise this goal, the policy suggests teaching students Mathematics and Computational
Thinking, along with new subjects like Artificial Intelligence, Machine Learning, and Data Science
during their school education. The policy also focuses on technology -enabled learning and
classrooms by using tools like artificial intelligence, machine learning, and adaptive testing to create
knowledge.

The National Curriculum for School Education draws from this policy aspiration and emphasizes the
need to introduce these emerging domains of study and technologies in the school curriculum. It
recommends inclusion of subjects such as design thinking, augmented reality, virtual reality, artificial
intelligence, and computational thinking. Additionally, it promotes the use of gamified content,
interactive content, and immersive experiences (such as AR, VR, or virtual labs) to enhance student
learning. In a variety of subjects, including design, music, art, and sciences, these resources support
students in knowledge creation and exploration, and development of capacities such as problem-
solving, critical and creative thinking.

CBSE, under the aegis of the Department of School Education and Literacy, Ministry of Education,
Govt. of India, is implementing a Curriculum on Computational Thinking and Artificial Intelligence (CT
& AI) to inculcate AI-readiness in school students. This curriculum will be implemented from classes
3rd to 8th, in the session 2026-27, and aims to develop AI-Ready learners, by focusing on
Computational Thinking Skills. The AI-readiness, so inculcated through CT Skills, will help develop
the capacities of learners to use computational thinking, such as logical thinking, problem solving,
pattern recognition, and so on, and understand the role and use of Artificial Intelligence in daily life.
The Curriculum aims to build strong foundations in computational thinking, digital literacy, and
responsible use of technology, along with nurturing innovation, critical thinking, and ethical decision-
making capacities.

Page 5

1. Relevance: Importance of Introducing CT and AI
Introducing CT and AI at the Grade 4 level is vital for positioning students as future-ready digital
citizens.
• Foundation for AI: Computational thinking is the intellectual backbone and cognitive framework
required to understand and eventually create AI-driven solutions
• Cognitive Development: It fosters essential human capacities such as logical thinking,
systematic problem-solving, and pattern recognition
• Preparation for the Future: Early exposure equips individuals with the ability to use data
effectively and apply technology ethically, which is necessary for the modern world of work
• Holistic Growth: It promotes interdisciplinary learning, helping students see that knowledge is
not compartmentalized by connecting Math, Science, and Humanities

2. Objectives (Curricular Goals)
For Grade 4 (part of the Preparatory Stage), the curriculum focuses on three primary goals:
• CG-1: Develop basic problem-solving skills with procedural fluency to solve daily-life problems
as a step toward formal computational thinking
• CG-2: Develop basic capacities of analytical thinking, verbal, and visual reasoning
• CG-3: Demonstrate understanding of basic concepts of computers and knowledge of hardware
and software

3. Learning Outcomes
ABSTRACT THINKING
Students will be able to solve moderate to highly moderate problems with partially visible or
incomplete ideas, using:
• Different viewpoints of 3D objects
• Changes in shapes after flips, turns, cuts/folds, or rotations
• Hidden or missing parts in incomplete shapes or patterns
• Mirror images and identical halves based on symmetry

PATTERN RECOGNITION
Students will be able to identify patterns involving one or more changes in consecutive terms,
formed using:
• Numbers
• Shapes or images
• Letters
• Or a mix of the above

DECOMPOSITION
Students will be able to break down problems involving a cluster of moderate clues, using
information from:
• Number clues (place values, sum/difference/product)
• 3D objects and their parts (faces, edges, corners)
• Step-by-step exchanges or transfers (money, objects, digits, quantities)
• Tables or charts with multiple pieces of information
• Conditions for counting/grouping/sorting items

Page 6

ALGORITHMIC THINKING
Students will be able to follow a set of well-defined, elaborate conditions to solve moderate to
complex problems involving:
• Number sequences formed using simple operations
• Movements on grids or direction-based paths
• Values that increase or decrease across steps
• Multi-step instructions involving moves, changes, transfers, swaps
• People/Events arranged in an order using attributes or chronological clues
• Simple counting instructions

4. Mapped with NEP and NCF 2023
The curriculum is directly aligned with national educational reforms:
• NEP 2020 Vision: It fulfils the goal of making India a global leader in emerging domains like AI
and Machine Learning by integrating them into school education
• NCF-SE 2023 Alignment: The learning standards (Goals Competencies Outcomes) are derived
from the framework suggested in the National Curriculum Framework for School Education 2023
• Phased Implementation: Following NCF recommendations, the curriculum introduces CT first
as a basis for learning AI later in higher classes

5. Time Allocation
• Annual Hours: A total of 50 hours annually is suggested for the Preparatory Stage (Classes 3–
5)
• Balanced Integration: To avoid overburdening students, this time is not added as an extra
subject but is integrated into existing Mathematics and "The World Around Us" (TWAU) periods

6. Approach / Pedagogy
The pedagogical approach for Grade 4 is designed to be playful and experiential:
• Activity-Based: Learning is driven by fun math games, puzzles, and hands-on exercises using
specialized worksheets
• Problem-Solving Focus: Teachers guide students to break larger problems into smaller parts
and interpret visual representations like charts and diagrams
• Collaborative Learning: The curriculum emphasizes peer discussions and group tasks to solve
problems collectively

7. Assessment
Assessment for Grade 4 shifts from rote memorization to continuous and competency-based
methods:
• Interactive Tools: Methods include written tests with CT puzzles, interactive group activities
and the use of a Teacher Observation Journal to track progress
• Qualitative Focus: The goal is to assess a student's ability to apply knowledge and think
creatively

Page 7

How to Use This Book?
This book is designed as a companion to the Mathematics textbook and is intended to be used
alongside regular classroom teaching. Since it follows the same chapter sequence, the
Mathematics teacher can seamlessly integrate it into daily instruction. As concepts are introduced
in class, the corresponding questions from this book can be used to deepen understanding and
encourage application.

Before beginning a chapter, the teacher is encouraged to go through the content of this book,
identify the underlying concepts required for each question, and plan how to align them with
classroom teaching. As these concepts are taught, the teacher can introduce the related thinking
questions to students.

It is important to note that the questions in this book are thinking-based and designed to promote
analysis, reasoning, and problem-solving. Teachers should adopt a facilitative approach, guiding
students through prompts and discussions rather than directly providing solutions. Students
should be given time to think and attempt independently, followed by classroom discussions
where different approaches are shared and explored.

Some chapters also include activities that build intuition and engagement. These should be
conducted before attempting the questions, as they help students approach the problems with
better understanding.

Teachers should approach this book with the mindset that the process of thinking is more
important than arriving at the correct answer. Creating a safe and encouraging environment where
students feel comfortable making mistakes, exploring multiple strategies, and expressing their
reasoning is essential. The goal is to nurture confident, independent thinkers rather than focus
solely on correctness.

Page 8

Chapter 1: Shapes Around Us
1. What will come in place of "?"
Note: You cannot rotate the images

a) b) c) d)
Answer: d
We have to subtract corresponding cubes, position-wise.
Figure out the blocks corresponding in both figures (check for the exact arrangement which is being
subtracted from the bigger arrangement):

We get,

Hence, we are left with:

Therefore, option d is correct.

Page 9

2. The figure below forms a cube when folded. Which of the following options represents a pair of
opposite faces of the cube?
Note: You cannot rotate the question or option images

a) b) c) d)
Answer: c
Let us first number the faces to avoid confusion.

Let’s identify the pairs of opposite faces.

As shown above, edges of the same colour are joined together to give the cube shape to the net.
The faces 1, 2, 4, and 6 are made to stand upright to form the cube.
Then, face 5 closes as a lid on the top to complete the cube shape.
Hence, the pairs of opposite faces in the final cube are:
1 and 6
2 and 4
3 and 5
Now, let’s analyse each of the following options.
Option a:

Faces 1 and 5 are not opposite. Hence, option a is invalid.

Page 10

Option b:

Faces 2 and 5 are not opposite. Hence, option b is invalid.
Option c:

4 and 2 are opposite faces. Hence, option c is correct.
Option d:

Faces 6 and 5 are not opposite. Hence, option d is invalid.
Hence, option c is the correct answer.
_____________________________________________________________________________________
3. From a rectangular piece of paper, a shape having which of the following number of angles
CANNOT be formed by making only a SINGLE STRAIGHT CUT in any direction?
a) 3 angles b) 4 angles c) 5 angles d) 6 angles
Answer: d
When we cut the rectangular piece of paper in a single straight cut, the following shapes can be formed:

Shape 1 has 3 angles after cutting.
Shape 2 has 4 angles after cutting.
Shape 3 has 5 angles after cutting.

A shape having 6 angles cannot be formed after cutting. Hence, the correct answer is option d.
_____________________________________________________________________________________
4. Count the number of TRIANGLES in the image given below.

a) 4 b) 5 c) 3 d) 2
Answer: a

Page 11

As highlighted in the image below, the figure contains 4 TRIANGLES.
Hence, the correct answer is option a.

_____________________________________________________________________________________
5. Find the ODD one out.

a) b) c) d)
Answer: a
The letter inside every shape is the first letter of its spelling and the number inside every shape is the
number of sides of that shape. Only option a does not follow this rule. Hence, the correct answer is option
a.
_____________________________________________________________________________________
6. If each of the given terms follows the same theme, what will come in place of "?"

a) b) c) d)
Answer: b
In each row, the total number of sides of the two yellow shapes gives the number of sides of the shape
in the middle. In the third row, there is a square and a parallelogram.
4 + 4 = 8. Option b has 8 sides. Hence, it is the answer.
_____________________________________________________________________________________
7. Each face of a triangular prism is coloured with a colour in such a way that no two adjacent faces
have the same colour. What is the MINIMUM number of colours required to do this?
Note: Two faces are said to be adjacent if they share a common edge
a) 2 b) 3 c) 4 d) 5
Answer: c
As shown in the image, there are 3 faces in the prism which would be adjacent to the front triangle face
(shown in yellow).

Page 12

So, all these four faces (including the triangle face) need to be coloured with different colours. The two
triangle faces are not adjacent to each other. So, they’ll be coloured with the same colour (yellow).
Hence, the MINIMUM number of colours needed to satisfy the given condition is 4.
So, the correct answer is option c.

_____________________________________________________________________________________
8. Which of these shapes will come in place of A and B, such that no two shapes having the same
number of sides, pattern, or colour appear in the same row or column?
Note: Not all blocks will necessarily have a shape present in them

a) b)

c) d)
Answer: a
Let’s begin with determining the shapes that can replace A and B, where we need to follow these
conditions:
• The shape for A should not have the same number of sides, pattern, or colour as the shapes in the
same row or column.
• Similarly, the shape for B should not have the same number of sides, pattern, or colour as the shapes
in the same row or column.
Option b is eliminated because the same pattern is present inside a shape in the same column as A and
shape of B is same as the pentagon in its row.

Option c is eliminated because shapes in white colour are present in the same row and column as A.

Option d is eliminated because the same shape is present in the same row as B.

Shapes in option a satisfy all the conditions. Hence, the correct answer is option a.

Page 13

9. Six concrete blocks of cubical shape are used for building the stairs as shown below. How many
faces of these blocks are not visible?

a) 12 b) 15 c) 21 d) 24
Answer: c
There are 6 cubical blocks in the figure.
Since a cube has 6 faces, the total number of faces would be: 6 × 6 = 36.
Since in the figure we can see a total of 15 faces only, the number of hidden faces would be: 36 - 15 =
21.
OR
Let us divide the figure into two portions - left side and right side.
Left side: 2 faces of each of the blocks are visible. For 3 cubes, we can see only 6 out of the 18 faces.
Therefore, 12 faces are hidden.
Right side: 3 faces of each of the blocks are visible. For 3 cubes, we can see only 9 out of the 18 faces.
Therefore, 9 faces are hidden.
Therefore, 21 faces (12+9) are hidden.
Option c is correct.
_____________________________________________________________________________________
10. How many edges of the pyramid shown below have a yellow dot at one of their corners but do not
have a red dot on the other corner?

a) 1 b) 2 c) 3 d) 4

Answer: b
As highlighted below, two edges of the pyramid have a yellow dot at one of their corners but do not have
a red dot on the other corner. Hence, the correct answer is option b.

Page 14

All the shapes of Set A must be placed into the squares present in Box 1, such that
• Each square has EXACTLY ONE shape
• The shape and the square CANNOT be of the same colour
• Two squares connected by a line CANNOT have the same shape inside them
Which option shows the shapes which are NOT next to each other in the final
arrangement?

(a) (b) (c) (d)
Answer: a
We need to place the shapes from Set A according to the given conditions. Each square can
hold only one shape, and the shape cannot be of the same colour as the square. Additionally,
the squares that are connected cannot contain the same shape. Following these rules, let's
begin placing the shapes from Set A into Box 1. For this, we have to logically identify the starting
step, a step in which we can accommodate a maximum number of shapes.
Consider this white square. It can either have a black triangle (case 1) or a black circle (case 2).

If this white square has a black triangle, the two black squares that are connected to it must
have two white circles. (As the black squares cannot have a triangle again) However, this case
is not possible, as our Set A does not have two white circles.

Hence, the white square has a black circle, and the black squares have white triangles.

Now, let’s consider this connected pair. Since connected squares must have different shapes,
the only white shape left, the white circle, goes in the black square. Therefore, the white square
must have a black triangle. (As a black circle cannot be placed here).

Finally, the black circle which is left in Set A must be placed in the empty white square.

Clearly, the black triangle and the black circle are NOT next to each other.
So, option a is the correct answer.

Page 15

Chapter 2: Hide and Seek
1. Sam travels from Start to End by visiting every block. He collects an item from every block except
from the block which comes immediately after a GREY BLOCK.
What is the TOTAL NUMBER of items he will collect?

a) 4 b) 5 c) 6 d) 7
Answer: b
As mentioned in the question, Sam cannot collect an item from any block that comes immediately after
a GREY BLOCK. Therefore, we first identify and cross out all such blocks.

Now, let's mark the remaining blocks from which Sam can collect the items.

Hence, Sam will collect a total of 5 items. Therefore, the correct answer is option b.
_____________________________________________________________________________________
2. The star must follow the arrows and return back to the square. How many different paths can it
take?

a) 4 b) 5 c) 2 d) 3
Answer: a
As highlighted below, there are a total of 4 different paths. Hence, the correct answer is option a.

Page 16

3. A, B, C, and D each go to one of the following places: the Zoo, the Shop, the Cafe, and the Gym.
● Each person visits a DIFFERENT location and starts from a DIFFERENT white square
● B reaches the Gym in exactly 1 turn
● A reaches the Cafe in exactly 2 turns
● D did not take any turns to reach his destination
Who among them started from the square having the question mark?

a) A b) B c) C d) D
Answer: c
Let us first label the empty boxes as 1, 2, 3, and 4.

B reaches the Gym in exactly one turn. 1 and 3 are the positions from which B can reach the Gym in
exactly one turn.

A reaches the Cafe in exactly two turns. Thus, A is at position 3. So, B has to be at 1.

The remaining destinations are Shop and Zoo, and the remaining positions are 2 and 4.
A person starting from 2 needs to take turns to go to any destination, while a person starting from 4 can
go straight to the Shop without taking any turns.
Since D did not take any turns to reach his destination, D must be at 4, and C must be at 2 as it is the
only position remaining.

Hence, the correct answer is option c.

Page 17

4. Sam wants to reach HOME by walking along the GREY STRIP. Identify the missing part of the
strip.
Note: You are not allowed to rotate any of the option images

a) b) c) d)
Answer: d
Let’s solve this logically.
As ‘HOME’ is at the bottom of the grid, we need to choose a path that ends at the bottom portion of the
grid.
Based on this, options a and b can be eliminated, as their paths end upwards and sideways, as shown
below:

Now, examine option c.

This path might take Sam to the bottom portion, but it does not have sufficient grey cells to make him
reach ‘HOME’, as shown below:

As we can see in the given image, option d will allow Sam to reach ‘HOME’. Hence, the correct answer
is option d.

Page 18

5. In the grid given below, which letter has the MOST number of arrows pointing towards it?

a) A b) B c) C d) D
Answer: b
Let’s check each of the options one by one and see which letter has the highest number of arrows pointing
towards it.
Option a: As shown in Image 1, A has 1 arrow pointing towards it.
Option b: As shown in Image 2, B has 3 arrows pointing towards it.
Option c: As shown in Image 3, C has 2 arrows pointing towards it.
Option d: As shown in Image 4, D has 2 arrows pointing towards it.

Since B has 3 arrows pointing towards it, which is the highest number of arrows pointing towards any
letter, option b is the correct answer.
_____________________________________________________________________________________
6. Which of the following options is the top view of the question image?
Note: You cannot rotate the question or option images

a) b) c) d)
Answer: c

As shown in the image, four blocks are visible when viewed from the top.
In this view, three blocks are arranged in a straight line, and the fourth block is positioned below the
middle one, forming a T-shaped pattern.
Hence, the correct top view is option c.

Page 19

7. Which of the views given in the options shows the highest number of different colours?

a) Front View – ARRANGEMENT 1 b) Right-side View – ARRANGEMENT 1
c) Front View – ARRANGEMENT 2 d) Right-side View – ARRANGEMENT 2
Answer: d
Let’s check each of the given options:
Option a) Front View – ARRANGEMENT 1:
When ARRANGEMENT 1 is viewed from the front, Red, Green, and Yellow, are visible.

Option b) Right-side View – ARRANGEMENT 1:
When ARRANGEMENT 1 is viewed from the right, Red and Blue are visible.

Option c) Front View – ARRANGEMENT 2:
When ARRANGEMENT 2 is viewed from the front, Red, Green, and Blue, are visible.

Option d) Right-side View – ARRANGEMENT 2
When ARRANGEMENT 2 is viewed from the right, Red, Yellow, Green, and Blue, are visible.

Since 4 different colours are visible, the correct answer is option d.

Page 20

8. Which of the following options will look like Figure X when viewed along the direction of the
arrow? Note: You cannot rotate the question or option images

a) b)

c) d)
Answer: c
We will analyse the view of each option from the given direction and then see if it matches with the Figure
X.
Option a: The right-side view doesn’t match Figure X.

Option b: The right-side view doesn’t match Figure X.

Option c: The right-side view is exactly the same as Figure X.

Option d: The right-side view doesn’t match Figure X.

Thus, option c is the only option whose view from the given direction matches Figure X.
Hence, the correct answer is option c.

Page 21

9. If a person is standing on block number 3, facing the animals, what will be the position of the cat
with respect to the fox from his point of view?

a) In front b) Behind c) Left d) Right
Answer: b
According to the image, if the person is standing on block number 3, the cat will be behind the fox.
Option b is correct.

_____________________________________________________________________________________
10. Which of the following figures is the top view of the solid?
Note: You cannot rotate the question or option images

a) b) c) d)
Answer: a
Top view is the view you get when you see any solid from the top.
If you see the solid from the top, you will be able to see only the blue faces as highlighted in the image
below. Thus, option a is the right answer.

Page 22

1. Ten coins are distributed among four people P, Q, R, and S such that one of
them gets one coin, another gets two coins, the third gets three coins, and
the fourth gets four coins. It is known that Q gets more coins than P; and S
gets fewer coins than R.
1. If R gets at least two more coins than S, then which one of the following is
necessarily true?
(a) Q gets at least two more coins than S
(b) Q and R together get seven coins every time
(c) P gets more coins than S
(d) P and Q together get at least five coins
Answer: b
From the question we can deduce that Q > P and R > S.
Now, if R gets at least 2 more than S, then the combination of coins distributed to
S and R could either be (1,3), or (1,4), or (2,4).
Case 1: S = 1, R = 3, P = 2, and Q = 4
Case 2: S = 1, R = 4, P = 2, and Q = 3
Case 3: S = 2, R = 4, P = 1, and Q = 3
In every case, Q and R together always have 7 coins.
Thus, option b is the correct answer.

Page 23

Chapter 3: Patterns Around Us
Activity Time
Odd and Even Numbers
Activity: Coin Flip Magic
Introduction
Using the concept of parity, which refers to the evenness or oddness of a number, we can identify the
change in certain patterns. These methods are used to analyse errors in our communication systems.
Here, we will explore an activity in which we play with two coloured coins (Black and White) arranged in
a grid to mimic data. We will flip a coin to simulate an error in the data. The activity describes a procedure
for precisely detecting errors in the grid pattern.

Activity Time Description

Launch 5 Teacher demonstrate the activity
min
Supporting Links:
Activity Video Link: https://youtu.be/-
t9VrqBOmz8?si=MGs2XOpU2hr035Zs
Reference: https://www.csunplugged.org/en/topics/error-
detection-and-correction/

Trial by 15 Student’s Tryout the activity with each other.
Students min
Student Worksheet:
https://docs.google.com/document/d/1nbcoVKiETEe-
4ta4zeppEcVCCs3NhznYJuWEvhmaFy4/edit?usp=sharing

Discussions 15 Attempting the worksheets and the discussion based on the
and min activity
Explorations

Page 24

CT Components
Algorithmic Thinking:
When we count the number of black coins in each row, one by one, and check whether the number is
even or odd, and we do the same thing for every column. Repeating these steps for every row and column
follows a clear algorithm to pinpoint the flipped coin.
Decomposition:
To identify the flipped coin, we break it down into steps: focus on one row, ask whether it has an even or
odd number of black cards, and do the same for every column.

Generalisation:
Once students understand the algorithm for identifying flipped coins, they will be able to find a flipped
coin on any size grid.

Logic:
By flipping any coloured coin, an even number of black coins in a row/column changes to an odd number.

Abstraction:
Grid patterns are a representation of data, and error in the data is represented by flipping the coin.

Activity: Coin Flip Magic
You have a grid and double-coloured coins with black and white sides. Ask someone to randomly arrange
the coins, white side up or black side up, in a 4×4 grid pattern. This will ensure that you are not familiar
with the pattern. You will add one more row and one more column, creating a 5×5 grid pattern with the
following details in mind. While adding the extra row and column, ensure each row and column has an
even number of black sides up (zero is also even). Close your eyes and have someone flip any one coin
on the grid pattern. When you open your eyes, you can figure out which coin was flipped. Let us see
how?

Figure: Activity Setup and Trick

1. When we flipped the black coin in the 5th row and 2nd column, what changes in the given row
and column?
a) The number of white coins increases, and it becomes odd
b) The number of black coins decreases, and it becomes odd
c) The number of black coins increases, and it becomes odd
d) The number of white coins decreases, and it becomes odd
Answer: b
You will notice that when you flip a coin with a black side up, the number of black sides up in a given row
and column decreases, and it becomes odd.

Page 25

By observing rows and columns that have an odd number of black coins, we can figure out which coin
was flipped in the activity.
Competency: :
Decomposition
Logic
_____________________________________________________________________________________
Explorations
1. Based on what you saw above, can you identify which coin was flipped in the given grid pattern?

a) b) c) d)
Answer: b
Observe that in the highlighted 2nd row and 4th column, the number of black coins is odd in number. By
observing rows and columns that have an odd number of black coins, we can figure out which coin was
flipped in the activity.

Competency:
Decomposition
Logic

Page 26

2. For the given 3 x 3 grid pattern, which would be the corresponding 4 x 4 grid pattern that will help
to detect which coin was flipped?

a) b) c) d)
Answer: c
In the grid pattern, observe that in all rows and columns, the number of black coins is even in number.

Competency:
Generalisation
Decomposition
Logic
_____________________________________________________________________________________
3. Instead of a 4×4 grid pattern, if we started with a 6×6 grid pattern, how many coins would be
needed for the extra row and column?
a) 13 b) 14 c) 15 d) 16
Answer: a

Competency:
Generalisation
Decomposition
Logic

The Flip "trick" is really not a magic trick at all. It is a precise algorithm based on the concept of parity.
Parity refers to the evenness or oddness of a number.

Remember, while adding the extra row and column, we ensured that each row and column has an even
number of black coins. These extra coins are called parity coins. These parity coins help us track the
“evenness” of the black coins before the flip and the change in “evenness” after the flip. This tracking is
done by identifying the row and column which has an odd number of black coins.

Page 27

Questions
1. Given below are 8 squares, to be filled with unique numbers from 1 to 8, and some numbers are
already filled as shown below.
If the result of all the four operations is always a single-digit ODD NUMBER, then which of the
following can NEVER be the result of any operation?

a) 1 b) 3 c) 5 d) 7
Answer: c
Some numbers are already placed in the grid, while the remaining numbers to be filled are 1, 4, 5, and
6. Since the result must always be a single-digit odd number, each odd number should be paired with an
even number and vice versa.
In the given squares, 3 and 7 (odd numbers) are already present. From the available numbers, the even
numbers 4 and 6 can be placed with 3 and 7 as shown below.

Similarly, 8 and 2 (even numbers) are already in the grid, and the remaining odd numbers, 1 and 5, can
be paired with them accordingly as shown below.

After placing all the numbers following this pattern, 5 never appears as a result in any case. Hence, the
correct answer is 5. Hence, the correct answer is option c.
_____________________________________________________________________________________
2. XY is a two-digit odd number. The sum of X and Y is a single-digit odd number. At maximum, how
many different possible values can X have?
a) 2 b) 3 c) 4 d) More than 4
Answer: c
XY is a two-digit odd number.
Y is definitely an odd digit.
Now, we don’t know whether X is even or odd.

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But it is given that the sum of X and Y is a single digit odd number.
If Y is an odd number and the sum of X and Y is a single digit odd number, then X should be an EVEN
number. (Since Odd digit + Even digit gives an odd number)
X can be 2, 4, 6 or 8. (X cannot be 0, as XY wouldn’t be a 2-digit number)
Therefore, X has 4 different possible values.
Thus, option c is the correct answer.
_____________________________________________________________________________________
3. Six consecutive even numbers are written on the faces of a single dice. If the smallest number is
8, then what will be the largest number?
a) 18 b) 28 c) 16 d) 20
Answer: a
Six consecutive even numbers starting from 8 are 8, 10, 12, 14, 16, and 18.
The largest number will be 18.
Thus, option a is the correct answer.
_____________________________________________________________________________________
4. What will come in place of "?" in the given series?

a) b) c) d)
Answer: d
As shown in the figure below:
The odd positioned terms are consecutive numbers starting from 2. The even positioned terms are
consecutive numbers starting from 4. So, the next term will contain 7 dots.

Hence, the correct answer is option d.

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5. What will come in place of "?"

a) b) c) d)
Answer: c
Even numbers on the left-hand side are replaced with a black coloured circle on the right side.
Based on this rule, option c is the correct answer.
_____________________________________________________________________________________
6. Diya wants to pick only the items placed on the steps marked with even numbers. Which of the
following options shows the food items she would collect?

a) 2 Pizzas, 2 Ice-creams, and 1 Apple b) 1 Pizza, 1 Ice-cream, and 2 Apples
c) 1 Pizza, 2 Ice-creams, and 1 Apple d) 2 Pizzas, 1 Ice-cream, and 1 Apple
Answer: c
Out of the given numbers, even numbers are 6, 4, 8, and 2. Hence, Diya will pick the following:
6 = Pizza
4 = Ice-cream
8 = Apple
2 = Ice-cream
i.e. 1 Pizza, 2 Ice-creams, and 1 Apple.
Hence, the correct answer is option c.

Page 30

7. Mani has some sweets with him. He wants to divide them between him and his friend Swami
equally. He tries many times but is not able to divide the sweets equally. What could be the
reason?
Note: A full sweet cannot be split into parts
a) The number of sweets that Mani has is even b) The number of sweets that Mani has is odd
c) The number of sweets that Mani has is 46 d) None of these
Answer: b
If Mani tries many times but still cannot divide the sweets equally between the two of them, it means the
number of sweets must be odd. An odd number cannot be split into two equal whole numbers.
So, the correct answer is option b.
_____________________________________________________________________________________
8. Statement: The sum of two natural numbers is 39.
Which of the given options can be understood from the given statement?
a) One of the two natural numbers is odd b) Both the natural numbers are odd
c) One of the two natural numbers is even d) Both a and c
Answer: d
The sum of two natural numbers can be odd only if one of them is even and the other is odd. Hence, both
first and third options can be understood. Hence, the correct answer is option d.
_____________________________________________________________________________________
9. There are only 4 people, A, B, C, and D, standing in a queue.
1. A is not standing at an even-numbered position from the front
2. D is standing immediately ahead of B
If B is not standing last, then who is standing last?
a) A b) B c) C d) D
Answer: c
There are only 4 people, A, B, C and D, standing in a queue. Let us have 4 positions from the front to the
back of the queue.

A is not standing at even numbered places from the front.
So, A must be standing either at positions 1 or 3.

D is standing immediately ahead of B.
So, Case 1 will have 2 different possible positions for D and B.
Case 2 has only one possible placement for D and B.
All of them are shown as follows:

Page 31

But B is not standing at the last position. So, Case 1 - b is invalid.
Clearly, in the remaining cases, we can see that C is standing at the last position.

Hence, option c is the correct answer.
_____________________________________________________________________________________
10. If any 4 odd numbers in the figure shown below are connected in the ascending order by using
straight lines, then which of the following figures CANNOT be made?

a) b) c) d)
Answer: d
As shown below, option d cannot be formed as it does not connect the dots in the ascending order.
Hence, the correct answer is option d.

Page 32

11. Question: Is M = N?
Statement 1: N is an even number.
Statement 2: M is an odd number.
To answer the given question, which of the given statements is/are sufficient?
a) Only 1
b) Only 2
c) Both 1 and 2 together
d) Question cannot be answered even if both statements are used
Answer: c
From statement 1, we know that N is an even number. So, statement 1 alone is insufficient.
From statement 2, we know that M is an odd number. So, statement 2 alone is insufficient.
Now, if we use both statements together, N is even and M is odd, which means they can never be
equal. So, the question can be answered if we use both the statements together.
Hence, option c is the correct answer.

Page 33

2. Given below is an arrangement of 6 circles, each filled with a different colour: Yellow, Black,
Purple, Red, Blue, and Green.
• The Yellow circle is connected to all circles except the Black circle
• The Black circle is only connected to the Purple circle
• The Yellow circle is the only circle that is connected to the Red circle
• The Blue circle is directly connected to the Yellow circle and to only one other circle
What is the colour of the circle marked with "?"

(a) Red (b) Purple (c) Green (d) Blue
Answer: c
The correct approach to solve this question is to analyse each statement one by one and assign the
circles accordingly.
• The Yellow circle is connected to all circles except the Black circle
• To identify the correct circle, we need to find the one that is connected to all others except one
• Once identified, we can highlight it and proceed with the remaining connections

The Black circle is only connected to the Purple circle. So, we need to find the circle that is connected
to the Black circle and assign it accordingly.

The Yellow circle is the only circle connected to the Red circle. Now, we need to identify a circle that
is only connected to the Yellow circle and assign it accordingly.

The Blue circle is directly connected to the Yellow circle and only one other circle. Now, among the
remaining two circles, we need to select the one that is connected to exactly two circles, with one of
them being the Yellow circle.

Hence, the colour of the circle which is marked with "?" is Green.
Hence, the correct answer is option c.

Page 34

Chapter 4: Thousands Around Us
Activity: Sorting Network
Introduction
Ever wondered how computers sort random things into order so quickly? Books in library, dictionaries,
list of students in the classroom, are arranged in an order for convenience. Without sorting, finding
information would be much more difficult. This activity uses the network template to be drawn on a floor
using chalk (or use flex print) and demonstrates how the numbers can be arranged in order.

Activity Time Description

Launch 15-20 Teacher will take the students to outdoor and conduct the
mins activity.

Supporting Links:
• Network template -
https://tinyurl.com/SortingNetworkCT
• Activity Video Link -
https://tinyurl.com/SortingNetworkVideo
• Reference -
https://www.csunplugged.org/en/topics/sorting-
networks/

Page 35

Trials by 10 Student’s try out the activity within a batch of 6. Others can
Students min observe the activity and take turns to try out.

Discussions 15 Attempting the worksheets and the discussion based on the
and min activity
Explorations

Computational Thinking Components
• Sorting Algorithm: The network sorts any given set of numbers in an ordered list. The design
of the network and related tasks provides opportunity to think about the logic and algorithm
behind the activity.
• Abstraction: The design of the network is an abstract representation of how actually data is
organised into computer.
• Decomposition: The network can be started with comparing two numbers and repeating that
basic comparison unit, we can sort any set of input numbers.

Explorations
Explore the activity with different orders of the numbers in the starting position. Are numbers getting
sorted at the end for every input sequence? Instead of numbers 1 to 6, take 2-digit and 3-digit numbers.
For example: {23, 56, 78, 113, 378, 667}.
Similarly, the same activity can also be done for sorting students with their names in an alphabetical
order. Students will ask each other’s name in the comparison box. The one with prior in order will go to
left, the other one to the right. See if the names are in order at the end or not at the end. Try the same
activity by comparing their height.
Consider the network to sort any 4 input numbers. The input numbers are {78, 56, 23, 113}. From the
input circle position, move the numbers forward to the rectangle box and compare them. The smaller
number will follow the blue line and bigger will follow the orange line.

Page 36

1. What will be correct positions of input numbers after the first level of comparison?

a)

b)
Answer: a
The smaller number moves along the blue line.
_____________________________________________________________________________________
2. What will be correct positions of input numbers after the second level of comparison?

a)

b)
Answer: b

Page 37

By following orange line from the upper box, 56 will reach the comparison box. By following the blue line
from the lower box, 78, will reach the comparison box.
Now consider the reversal in the rule. Instead of the smaller number following the blue, now, the bigger
number from the comparison box will follow the blue arrow. The smaller number will follow the orange
arrow.
_____________________________________________________________________________________
3. What will be the output order of the numbers at the end, if the comparison box rule is reversed?
a) No difference, output will be same
b) Output numbers will be in reverse order - descending
c) The output depends on input number order
d) Can’t say anything about the order
Answer: b
At every comparison, the bigger number will follow the blue line and move upwards while smaller number
will follow the orange line and move downwards.

Page 38

Questions
1. 15A9 and 1A79 are two 4-digit numbers where A represents one of their digits. Which of these
conditions about these 4-digit numbers CAN be true, when A is less than 6?
a) 15A9 > 1A79 b) 15A9 < 1A79 c) 15A9 = 1A79 d) Either A or B
Answer: d
The numbers given are 15A9 and 1A79
Step 1: Compare place values
Both numbers are 4-digit numbers, so we compare the digits from left to right.
Thousands place: Both have 1 in the thousands place.
Step 2: Hundreds place: We have A and 5 in hundreds place.
A is less than 6.
A can be 5, 4, 3, 2, 1 or 0.
If A = 5, then 1559 < 1579.
If A < 5, say 4, then 1549 > 1479.
So, for A = 5, option b is true and for any other value less than 5, option a can be true.
Hence, option d is the correct answer.
_____________________________________________________________________________________
2. What is the smallest possible 4-digit number formed using 4 different circles from the grid given
below, such that the number of white circles in the number formed is less than that of the black
circles?

a) 1360 b) 1023 c) 1063 d) 1036
Answer: d
The number of black circles in the 4-digit number formed must be more than the white circles.
As the grid has only 3 black circles, the number of black and white circles used in the 4-digit number must
be 3 and 1 respectively.
The black circles contain the digits 6, 3, and 1.
Among the white circles, 0 must be used to minimize our number.
So, the smallest possible 4-digit number that can be formed is 1036.
Option d is correct.
_____________________________________________________________________________________
3. In the given expression, 56B1 and 5B60 represent two 4-digit numbers, where "B" represents one
of the digits. How many DIFFERENT VALUES can "B" take from SET B to satisfy the given
condition?

a) 5 b) 4 c) 3 d) 2
Answer: b
Step 1: Compare digits place by place
Thousands place: Both numbers have 5 thus they are equal.
Hundreds place: Compare 6 (in 56B1) with B (in 5B60).

Page 39

For 56B1 > 5B60, the hundreds digit of the first number (6) must be greater than the hundreds digit of
the second number (B).
That means:
6>B
If we put B as 6,
5661 > 5660, it also satisfies the condition.
Hence B can be less than or equal to 6.
There are 4 numbers in SET B which are less than or equal to 6
Therefore, option b is correct.
_____________________________________________________________________________________
4. Using the digits 4, 8, 7, 6, 1, 3, and 5, form the largest and the smallest 4-digit numbers such that
no digit repeats within a number. Which digit will be common in both numbers?
a) 3 b) 4 c) 5 d) 6
Answer: c
To create the largest and the smallest 4-digit numbers using the digits 4, 8, 7, 6, 1, 3, and 5 ensuring that
each digit is unique within a number, we can arrange them accordingly.
The largest 4-digit number:
Arrange the larger digits in descending order: 8765
The smallest 4-digit number:
Arrange the smaller digits in ascending order: 1345
Now, between 8765 and 1345, we can observe that both numbers share the digit 5.
Option c is correct.
_____________________________________________________________________________________
5. If you remove the digit that is common to all four numbers shown below, which number will
become the largest?
Note: You cannot re-arrange the digits to form a number

a) Number 1 b) Number 2 c) Number 3 d) Number 4
Answer: a
Step 1: Find the digit common to all four numbers
Check digits present in each:
49526: {4, 9, 5, 2, 6}
21724: {2, 1, 7, 2, 4}
68401: {6, 8, 4, 0, 1}
64851: {6, 4, 8, 5, 1}
The digit 4 appears in every number. So, we must remove 4 from each.
Step 2: Remove the digit 4 (without rearranging)
Number 1: 49526. Remove 4 = 9526
Number 2: 21724. Remove 4 = 2172
Number 3: 68401. Remove 4 = 6801
Number 4: 64851. Remove 4 = 6851
Step 3: Compare the resulting numbers
9526 (Number 1)
2172 (Number 2)
6801 (Number 3)

Page 40

6851 (Number 4)
The largest value is 9526 (Number 1).
Therefore, option a is correct.
_____________________________________________________________________________________
6. Which of the following numbers can come in place of "#" in the given expression?

a) 5321 b) 6489 c) 4897 d) 4678
Answer: d
Write down the condition clearly
5678 > # < 4687
This means ‘#’ must be smaller than both 5678 and 4687.
So, # < 4687 is the stronger condition to satisfy.
From the given options, only 4678 satisfies both inequalities.
Therefore, option d is correct.
_____________________________________________________________________________________
7. How many numbers from Set A can be placed in the blank, to form a number greater than 6632?

a) 1 b) 2 c) 3 d) 4
Answer: b
Given: Number formed = 6_44
Set A = {4, 5, 6, 7}
We need to find how many numbers from Set A can be placed in the blank so that the number formed is
greater than 6632.
Numbers greater than 6632: 6644 and 6744
Therefore, option b is correct.
_____________________________________________________________________________________
8. If you add ALL the digits in the HUNDREDS place of the numbers given below, what will be the
digit in the TENS place of the resulting number?

a) 1 b) 2 c) 3 d) 0
Answer: b
Step 1: Identify the digits in the HUNDREDS place of the given numbers
In 5432, hundreds digit = 4
In 2879, hundreds digit = 8
In 1945, hundreds digit = 9
Step 2: Add these digits

Page 41

4 + 8 + 9 = 21
Step 3: Find the digit in the TENS place of 21
The tens digit in 21 is 2.
Therefore, option b is correct.
_____________________________________________________________________________________
9. How many numbers in Box A, after swapping their tens and thousands digits, will be less than
3000?

a) 4 b) 3 c) 2 d) 1
Answer: c
Identify the thousands and tens digits.
Let’s write each number in the form:
(Thousands) (Hundreds) (Tens) (Ones)

Swap the thousands and tens digits.
We interchange the 1st (thousands) and 3rd (tens) digits in each number.

Check which numbers are less than 3000.
Now we compare each new number with 3000:
2435 < 3000
4657 > 3000
2473 < 3000
5881 > 3000
There are 2 numbers (2435 and 2473) that are less than 3000.
Therefore, option c is correct.
_____________________________________________________________________________________
10. How many numbers from "Image A" can be placed in BOTH the boxes given below?

a) 2 b) 3 c) 4 d) 5
Answer: b
Step 1: Filter for numbers greater than 4550
Compare each number with 4550:

Page 42

4856 > 4550, keep
4674 > 4550, keep
4775 > 4550, keep
4326 < 4550, exclude
4436 < 4550, exclude
4563 > 4550, keep
4653 > 4550, keep
4532 < 4550, exclude
So, the numbers remaining after this filter are: 4856, 4674, 4775, 4563, and 4653.
Step 2: From these, keep only odd numbers
A number is odd if its last digit is odd (1,3,5,7,9).
Check the last digit of each number:
4856: last digit 6 is even. exclude
4674: last digit 4 is even. exclude
4775: last digit 5 is odd. keep
4563: last digit 3 is odd. keep
4653: last digit 3 is odd. keep
The numbers that satisfy both conditions are: 4775, 4563, and 4653.
Therefore, option b is correct.

Page 43

A, B, and C want to meet. They cannot meet in the boxes they are currently
standing in or in any boxes immediately connected to them. In how many
boxes can they meet?

(a) 1 (b) 2 (c) 3 (d) 4
Answer: b
A, B, and C cannot meet in any of the boxes that are immediately connected to
them.
So, the first step is to cancel out the boxes that are immediately next to A.

Next, cancel out the boxes that are immediately next to B.

Now, cancel out the boxes that are immediately next to C.

So, there are 2 boxes where A, B, and C can meet, as highlighted below:

Hence, the correct answer is option b.

Page 44

Chapter 5: Sharing and Measuring
1. How many bottles in the image below are filled to less than half their capacity?

a) 2 b) 3 c) 4 d) 5
Answer: c
As shown below, there are 4 bottles in which less than half of the portion is filled. Hence, the correct
answer is option c.

_____________________________________________________________________________________
2. What fraction of the given image is shaded grey?

a) 5/18 b) 5/9 c) 6/18 d) 7/9
Answer: c
As shown in the figure below, if we divide each square into 2 triangles, then there are altogether 18
triangles, out of which 6 are shaded grey. Thus, 6/18 of the shape is shaded grey.
Hence, the correct answer is option c.

_____________________________________________________________________________________
3. Diksha bought some pens for exactly Rs. 80. If each pen costs the same, which of the following
CANNOT be the price of a pen?
a) Rs. 4 b) Rs. 5 c) Rs. 6 d) Rs. 8
Answer: c

Page 45

Diksha bought some pens for Rs. 80. The price of all the pens is equal.
Option a: 80÷4 = 20 (20 pens can be bought)
Option b: 80÷5 = 16 (16 pens can be bought)
Option c: 80÷6 = 13.33 (we cannot buy pens in fractions)
Option d: 80÷8 = 10 (10 pens can be bought)
Hence, the price of a pen cannot be Rs. 6. Therefore, the correct answer is option c.
_____________________________________________________________________________________
4. What will come in place of "?" in the given series?

a) b) c) d)
Answer: a
The series follows two rules:
1. Each time, the smaller (inner) circle increases by a quarter portion (in anti-clockwise order).
2. The bigger (outer) circle decreases by a quarter portion (in clockwise order).
From the 1st to the 3rd figure, the inner circle increases from ¼ to ½ to ¾, while the outer circle loses
one quarter each time.
So, in the next figure:
• The inner circle becomes complete
• The outer circle appears as a quarter
Only option a matches this.
Hence, the correct answer is option a.
_____________________________________________________________________________________
5. Which of the following options shows the correct number of minutes in (1/4 + 1/2) hour?
a) 5 × 8 minutes b) 60 - 10 minutes c) 21 + 24 minutes d) 3 × 10 minutes
Answer: c
We know that 1 hour = 60 minutes.
So, 1/4 of an hour = 60 ÷ 4 = 15 minutes and
1/2 of an hour = 30 minutes
So, 30 + 15 = 45 minutes
Now check each option:
● 5 × 8 = 40 minutes
● 60 - 10 = 50 minutes
● 21 + 24 = 45 minutes. This is equal to (1/4 + 1/2) hour.
● 3 × 10 = 30 minutes
Therefore, option c is correct.

Page 46

6. Find the ODD one out.

a) b) c) d)
Answer: d
All the other options have half a pizza, except option d, which has three-quarters of a pizza.
Hence, the correct answer is option d.
_____________________________________________________________________________________
7. If each of the given terms follows the same theme, what will come in place of "?"

a) b) c) d) None of these
Answer: a
In each row, the left box contains two horizontal bars, a top bar and a bottom bar.
Each bar is divided into parts, some shaded and some white.
To decide the symbol on the right side, we must consider the total shaded and total white parts together
from both bars.
Observe the given rows:
First row:
The total shaded area (top + bottom) is equal to the total white area.
So, the symbol shown is “=”.
Based on the same rule, in the last row, when we count the top and bottom bars together, the total shaded
area and the total white area are equal.
So, the correct symbol should be “=”.
Hence, the correct answer is option a.

Page 47

8. What will come in place of "?"

a) b) c) d)
Answer: a
We observe that in each given figure, a fraction represents the number of sides of two different colours.
The numerator and denominator are also shown in the colour corresponding to the edges.
The smaller number is written as the numerator. For example, in the first term, one edge is red and the
other three are blue. Therefore, the resulting fraction is 1/3, where 1 is shown in red and 3 in blue.
Here, we have 2 edges in purple and 4 edges in pink.
Hence, the answer should be 2/4, where 2 must be in purple colour and 4 in pink colour.
Hence, option a is correct.
_____________________________________________________________________________________
9. A piece of paper is shaped like a rhombus, but not a square. At most, how many times can it be
folded so that each fold divides it into two identical halves?
a) 0 b) 1 c) 2 d) 3
Answer: c
The given paper is a rhombus, but not a square.
First fold:
The paper can be folded exactly into two equal parts along the diagonal BD. Both sides match perfectly.

Second fold:
The folded paper can again be folded into two equal parts along line OC, as shown in the figure.
After the second fold, the paper becomes a triangle with unequal sides.

In this shape:
- There is no way to fold the paper so that both parts match exactly
- No folding line divides the figure into two identical parts

Page 48

So, the paper cannot be folded into equal halves anymore.
The rhombus can be folded into identical halves at most 2 times.
Hence, the correct answer is option c.
_____________________________________________________________________________________
10. What will come in place of "?"

a) b) c) d)
Answer: b
Observed pattern
Number of parts:
The circle in the middle is divided into as many parts as the total number of sides of the two
shapes on its left and right.
Colours used:
The parts of the circle are coloured using the same two colours that appear on the adjacent
shapes.
Colour distribution (opposite mapping):
Within a row, the parts of the circle are filled using the opposite colour pattern of the two
adjacent shapes.
For example, in Row 1 the triangle has 3 blue sides, and the rectangle has 4 red sides.
So, the circle’s parts representing the triangle are coloured red, and the parts representing the
rectangle are coloured blue.
Applying the rule to Row 3
● The two shapes in Row 3 together have 8 sides in total
● Therefore, the circle should be divided into 8 parts
● Following the opposite-colour rule, 3 parts should be coloured purple, and 5 parts should
be coloured grey
Hence, the correct answer is option b.

Page 49

At a party, there are 10 people. Each person at the party knows at least one
other person there. Which of the following statements can never be true?
Note: If A knows B, then it does not necessarily mean that B knows A
(a) Every person knows only one other person
(b) Every person knows all the other people
(c) Every person knows a different number of people
(d) Every person knows exactly two other people
Answer: c
There are 10 people at the party.
Each person knows at least one other person. So, each person can know 1, 2, 3,
4, 5, 6, 7, 8, or 9 people. Now, let’s check the options.
Option a: Every person knows only one other person
This is possible. Each person can know just one person. So, this can happen.
Option b: Every person knows all the other people
Each person would know 9 people. This is also possible.
Option c: Every person knows a different number of people
The possible numbers are 1 to 9. But there are 10 people.
Since there are only 9 different numbers, two people must have the same
number.
So, it is not possible for all 10 people to have different numbers.
Option d: Every person knows exactly two other people
This is also possible.
Therefore, the statement that can never be true is: Every person knows a different
number of people.
Hence, the correct answer is option c.

Page 50

Chapter 6: Measuring Length
1. In the image given below, there are 20 small squares, each of the same size. Which line has the
longest part hidden behind the black squares?

a) b) c) d) Both a and b
Answer: c
The line in option a has two parts behind the black squares (refer to Image 1)
The line in option b also has two parts behind the black squares (refer to Image 2)
However, the line in option c has three parts behind the black squares (refer to Image 3)
Hence, the answer is option c.

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2. If Block __ is removed from the given Figure, the perimeter of the Figure will be decreased. Fill in
the blank appropriately.

a) Q b) S c) R d) P
Answer: d
If we remove Block Q, one side is removed, and 3 sides are added to the perimeter.
Therefore, the perimeter increases.

Similarly, on removing Block S, the perimeter increases.

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If we remove Block R, two sides are removed, and 2 sides are added. Therefore, the perimeter remains
the same.

If we remove Block P, three sides are removed, and one side is added. Therefore, the perimeter
decreases.

Hence, option d is the answer.
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3. Every coloured line represents a path and through this path the boy on the LEFT can reach the
RIGHT side. If he chooses the shortest path, how many junctions will he come across?
Note: A junction is the point where any two coloured lines cross each other

a) 6 b) 7 c) 8 d) 9
Answer: c
The shortest way is the black one. If he goes this way, he will come across 8 junctions.
Hence, the correct answer is option c.

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4. Count the number of blocks that have ALL THREE lines of different lengths.

a) 3 b) 4 c) 5 d) 6
Answer: b

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There are 4 blocks with ALL THREE lines of different lengths.

Hence, the correct answer is option b.
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5. Given below are lines of different lengths. They are to be categorized into three groups based on
their lengths:
• Group 1 for lines taller than line E
• Group 2 for lines shorter than line D
• Group 3 for lines taller than line B
Which line will belong to all three groups?

a) A b) F c) C d) D
Answer: a
Identifying lines based on their height:
Group 1 includes lines taller than E:
A, B, C, D, F, H
Group 2 consists of lines shorter than D:
A, B, E, G, H
Group 3 comprises lines taller than B:
A, C, D, F
Line A is present in all three groups. Thus, option a is the correct answer.
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6. In the set of bars given below, which bar does NOT have any other bar of the SAME height?

a) Bar E b) Bar D c) Bar C d) Bar F
Answer: d
Bars A and D are of the same height, as shown in the image.

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Bars B, C, and E are of the same height, as shown in the image below.

Only Bar F does not have any other bar of the same height. Option d is the correct answer.
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7. Which option will come in place of "?"

a) % + / > b) / % + > c) % > + / d) % + > /
Answer: d
Each bar has a symbol inside it. After the arrow, the symbols are arranged from the tallest bar to the
shortest bar.
The same rule is followed throughout.
Applying this rule to the last set, arranging the symbols from highest bar to lowest bar gives:
%+>/
Hence, option d is the correct answer.
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8. Each of the dogs A, B, and C use the red, yellow, and pink paths (in the same order), to go to their
bone. If each dog takes 10 minutes to reach the bone, which dog runs the fastest?

a) Dog A b) Dog B
c) Dog C d) Cannot be determined
Answer: c
Since each dog takes 10 minutes to reach the bone, we need to count the number of blocks each of them
is travelling. Therefore, Dog A has to travel 17 blocks, Dog B has to travel 15 blocks, and Dog C has to
travel 19 blocks. Since Dog C is able to cover more blocks in the same time, Dog C is the fastest.
Hence, the correct answer is option c.

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9. If the monkey follows the shortest path to reach the banana, which of the following options
represents the tiles that will appear immediately one after another on its path? The tiles must
appear in the same SEQUENCE as they appear in the original path.
Note: The monkey can move only horizontally or vertically along the tiles, not diagonally

a) b) c) d)
Answer: d
The shortest path to reach the banana is shown below.
As can be seen, the tiles in option d appear consecutively on the monkey's path.
Hence, option d is the correct answer.

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10. The cube shown below is cut horizontally along the red dotted line into two EQUAL halves. In
each resulting shape, how many faces will still have the same perimeter as a face of the original
cube?

a) 1 b) 2 c) 4 d) 6
Answer: b
Let the length of each face of the original cube be ‘a’.
We know that the face of a cube is in the form of a square.
So, the perimeter of each face of the original cube = a + a + a + a = 4a.
Now, the cube is cut into two halves horizontally.
This means that the height of the original cube decreases.
The height becomes a/2.

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The left, right, front, and back faces of the cube are affected due to this, but the top and bottom faces will
not have any change.
This means that the top and bottom faces of each resultant shape will still have the same perimeter 4a,
like in the original cube.
Hence, option b is the correct answer.
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E-1, E-2, and E-3 are three engineering students writing their assignments at
night. Each of them starts at a different time and completes the assignment
at a different time. The digit in their name cannot match the order in which
they start or complete the assignment. (For example, E-1 cannot be the first
to start or the first to complete the assignment because his name contains
the digit 1.) The last student to start is the first to complete the assignment.
Who is the first student to start writing the assignment?
(a) E-1 (b) E-2
(c) E-3 (d) Cannot be determined
Answer: c
The last to start is the first to complete and this cannot be either E-1 or E-3 as the
digit in their names is certainly not the same as the order of their starting or
ending the assignment. Thus, the last to start and the first to finish has to be E-2.
Now, the first to start cannot be E-1 as he has 1 in his name.
Therefore, E-1 could have only started second. Hence, E-3 started first.
Hence, option c is the correct answer.

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Chapter 7: The Cleanest Village
1. A and B have Rs. 200 each. Each of them visited a shop and bought an item. A spent Rs. 110 more
than B. What could be the highest possible amount spent by B?
a) Rs. 200 b) Rs. 180 c) Rs. 100 d) Rs. 90
Answer: d
Given that A and B have Rs. 200 each.
A spent Rs. 110 more than B.
The maximum amount that B can spend:
= 200 – 110 = 90.
If B spends more than Rs. 90, then, the amount spent by A would be 110 plus the amount spent by B.
This exceeds Rs. 200, which is contradictory, as A has Rs. 200 only.
Hence, option d, Rs. 90 is the highest possible amount spent by B.
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2. The monkey starts at 0 points and needs to climb the staircase to the END. On each step, it either
gains or loses points. Which ONE step must the monkey SKIP to earn exactly 114 points in total?

a) A b) B c) C d) D
Answer: b
The monkey starts with 0 points and must end with exactly 114 points.
There are four steps: A, B, C, and D. Each step either gives or takes away points.
The monkey must skip exactly ONE step.
Let’s check each possibility one by one.
Step A (+111 points)
If the monkey skips Step A, the remaining points would come from B, C, and D only.
But Step A contributes a positive value (+111).
Skipping it makes it impossible to reach a total as large as 114.
So, Step A cannot be skipped.

Step B (-110 points)
If the monkey skips Step B, the total is:
● Step A: +111
● Step C: +117
● Step D: -114

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Total = 111 + 117 - 114
Total = 114
This matches the required score.
Skipping Step B works.
Step C (+117 points)
If the monkey skips Step C, the total becomes:
● Step A: +111
● Step B: -110
● Step D: -114
Total = 111 - 110 - 114
Total = -113
This is not 114.
Skipping Step C does NOT work.
Step D (-114 points)
If the monkey skips Step D, the total becomes:
● Step A: +111
● Step B: -110
● Step C: +117
Total = 111 - 110 + 117
Total = 118
This is not 114.
Skipping Step D does NOT work.
The monkey must skip Step B to end with exactly 114 points.
Hence, the correct answer is option b.
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3. Boxes A, B, and C each have 500 chocolates. Diya removed 222 chocolates from box B and 222
chocolates from box C. Nihal added 111 chocolates to box B, and Jay removed 111 chocolates
from box A. Which boxes have an equal number of chocolates at the end?
a) A and B
b) A and C
c) B and C
d) No two boxes have the same number of chocolates
Answer: a
A, B, and C have 500 chocolates each.

Diya removed 222 chocolates from boxes B and C each.

Nihal added 111 chocolates to box B.

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Jay removed 111 chocolates from box A.

Now, box A and box B have 389 chocolates each and box C has 278 chocolates.
Hence, the boxes that have an equal number of chocolates in the end are box A and box B.
Therefore, the correct answer is option a.
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4. A, B, and C represent 3 different single-digit numbers in the grid given below. If each of these
options represents a 3-digit number formed using A, B, and C, find the option having the highest
value.

a) BAC b) BCA c) CBA d) ABC
Answer: b
First, let’s solve for A:
From the equation:
4+A=5
So, 4 + 1 = 5. Therefore, A = 1.
Now, let’s solve for C:
C+5=8
So, 3 + 5 = 8. Therefore, C = 3.
Next, solve for B using the equation:
7-B=C
7-B=3
So, 7 - 4 = 3. Therefore, B = 4.
Now, we know the values:
•A=1
•B=4
•C=3
Hence, the highest 3-digit number is: 431, which is BCA. Therefore, the correct answer is option b.
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5. Which of the following is the LEAST possible number that can replace the "?" such that the sum
of the numbers in Row 2 is GREATER than the sum of the numbers in Row 1?

a) 221 b) 222 c) 223 d) 224
Answer: d

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